Problem: $f(x) = \begin{cases} 5 & \text{if } x = 2 \\ 2x^{2}-1 & \text{otherwise} \end{cases}$ What is the range of $f(x)$ ?
Explanation: First consider the behavior for $x \ne 2$ Consider the range of $2x^{2}$ The range of $x^2$ is $\{\, y \mid y \ge 0 \,\}$ Multiplying by $2$ doesn't change the range. To get $2x^{2}-1$ , we subtract $1$ So the range becomes: $\{\, y \mid y ≥ -1 \,\}$ If $x = 2$, then $f(x) = 5$. Since $5 ≥ -1$, the range is still $\{\, y \mid y ≥ -1 \,\}$.